Discrete Mathematics, Spring 2004 Homework 9 Sample Solutions
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1 Discrete Mathematics, Spring 00 Homework 9 Sample Solutions. #3. A vertex v in a tree T is a center for T if the eccentricity of v is minimal; that is, if the maximum length of a simple path starting from v is less than or equal to the maximum length of a simple path starting from any other vertex w. Show that a tree has either one or two centers. Solution. Let P be a simple path of maximal length in T. (Such a path surely exists; in fact, we have seen this construction before, when establishing equivalent characterizations of trees.) The length of this path is either even or odd. In the first case, denote this length by m; then we have P = (v 0,v,...,v m,v m,v m+,...,v m,v m ), where all of the v i are distinct. We claim that v m is the unique center for T. To see this, notice first that the eccentricity of v m is actually equal to m (clearly it must be at least m). If it were not, then there would exist a simple path (v m = w 0,w,...,w k = w) with k > m. Since trees are acyclic, there are only two possibilities: w j = v m+j for all j some nonnegative integer j 0, and w j / {v 0,...,v m } for all j > j 0, or w j = v m j for all j j 0, and w j / {v 0,...,v m } for all j > j 0. In the first case, the path P = (v 0,v,...,v m = w 0,w,...,w k = w) would be a simple path of length greater than that of P, contradicting the way in which we chose P. In the second case, the path P = (w = w k,...,w,w 0 = v m,v m,...,v 0 ) would be a simple path of length greater than that of P, again a contradiction. Hence the eccentricity of v m is equal to m. Now consider any of the remaining vertices u in T. If u is one of the v i, then its eccentricity must be > m, for if u = v i with i m then (v i,v i+,...,v m ) is a simple path of length > m, and if u = v i with i > m then (v i,v i,...,v 0 ) is a simple path of length > m. If u is not one of the v i, then consider a simple path Q = (u = u 0,u,...,u r = v ), and let k 0 be the first index so that u k0 is one of the v i. Then at least one of the paths (u 0,u,...,u k0 = v i,v i+,...,v m ), (u 0,u,...,u k0 = v i,v i,...,v 0 ) will be a simple path of length m. We conclude that u has eccentricity greater than m, and hence v m must be the unique center of T.
2 In a similar fashion, one can show that if the length of a maximal simple path P is odd, say P = (v 0,...,v m ), then the vertices v m and v m are centers for T. Hence a tree has at most two centers. (Note: This proof is obviously fairly complicated; to see what s going on, it helps to draw a picture of the situation.). #. Draw a tree having four internal vertices and six terminal vertices, or explain why no such graph exists. Solution. A sample such tree is shown below:. #3. If a forest F consists of m trees and has n vertices, how many edges must F have? Solution. Let n i be the number of vertices in the ith tree. Then the ith tree has n i edges, and since m i= n i = n, the total number of edges is m (n i ) = m n i m = n m. i= i= i=
3 .3 #3. Use breadth-first search (Algorithm.3.) with the vertex ordering chbgadfe to find a spanning tree for the graph G given below: a e b e c e d e 3 e 0 e 9 e e e e e f e5 g e h Solution. (Note that replacing e 9 with e 0 in the graph below would also be valid.) a e b e c d e e 9 e e e f e 5 g h 3
4 .3 #. Use depth-first search (Algorithm.3.) with the vertex ordering dhcbefag to find a spanning tree for the graph G given in.3 #3. Solution. a e b c e d e e e e f e5 g e h.3 #. Write a depth-first search algorithm to test whether a graph is connected. Solution. We take Algorithm.3. and modify it just a little bit. The key point to observe is that no matter what graph G input into Algorithm.3., the tree T output will be the spanning tree for the component of G containing v. Hence if G is not connected, there should be less vertices in T than in G. The algorithm on the next page is a possible implementation. Observe that the input is the set V of vertices and the set E of edges of G, and the output is true if the graph is connected and false otherwise. During the course of the algorithm, T refers to the graph (V,E ).
5 procedure isconnected (V, E) V := {v } E := w := v done := false while done = false do while there is an edge (w,v) that when added to T does not create a cycle in T do choose such an edge (w,v k ) with minimum k add (w,v k ) to E add v k to V w := v k if w = v then else done := true w := parent of w in T if V = V then else return(true) return(false) end isconnected 5
6 . #5. Find the minimal spanning tree given by Prim s algorithm (Algorithm..3) for the graph below: Solution. A minimal spanning tree that results from applying Prim s Algorithm is shown below. The roman numerals refer to the order in which the edges were added. (i) 3 5 (vi) 9 (ii) (iv) (viii) 0 (iii) (v) (xiv) (vii) (xiii) (xii) (ix) (xv) (x) (xi) 3 5. #. Let G be a connected, weighted graph and let v be a vertex in G. Suppose that the weights of the edges incident on v are distinct. Let e be the edge of minimum weight incident on v. Must e be contained in every minimal spanning tree? The statement is true; we prove it by contradiction. Suppose the claim is false. Then there is a minimal spanning tree T for G which does not contain e. Let T be the graph obtained by adding the edge e to T; then T contains a cycle C, which in turn contains the
7 edge e. Clearly there must exist another edge e in C which is incident on v. Removing this edge, we then obtain another spanning tree T for G. However, calculating the weight of T, we find that weight(t ) = weight(t) + weight(e) weight(e ) < weight(t), where the inequality holds because e has less weight than any other edge incident on v. As T was assumed to be a minimal spanning tree, we have a contradiction, and the claim follows.. #5. Decide if the following statement is true or false. If true, prove it; otherwise, give a counterexample. If e is an edge in G whose weight is less than the weight of every other edge, e is in every minimal spanning tree of G. The statement is true; we prove it by contradiction (the proof is almost identical to the one given above, so I merely include it for completeness sake). Suppose the claim is false. Then there is a minimal spanning tree T for G which does not contain e. Let T be the graph obtained by adding the edge e to T; then T contains a cycle C, which in turn contains the edge e. Removing an edge e e, we then obtain another spanning tree T for G. However, calculating the weight of T, we find that weight(t ) = weight(t) + weight(e) weight(e ) < weight(t), where the inequality holds because e has less weight than any other edge in G. As T was assumed to be a minimal spanning tree, we have a contradiction, and the claim follows.
8 . #. Show how Kruskal s algorithm finds a minimal spanning tree for the graph given in. #5. Solution. A minimal spanning tree that results from applying Kruskal s Algorithm is shown below. The roman numerals refer to the order in which the edges were added. (xii) 3 5 (x) 9 (i) (viii) (xi) 0 (v) (xv) (iv) (iii) (xiv) (ix) (vii) (ii) (vi) 3 5 (xiii)
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